Calculate first the number of moles of ethylene glycol by dividing the mass by the molar mass.
n = (6.21 g ethylene glycol) / 62.1 g/mol
n = 0.1 mol
Then, calculate the molality by dividing the number of moles by the mass of water (in kg).
m = 0.1 mol/ (0.025 kg) = 4m
Then, use the equation,
Tb,f = Tb,i + (kb)(m)
Substituting the known values,
Tb,f = 100°C + (0.512°C.kg/mol)(4 mol/kg)
<em>Tb,f = 102.048°C</em>
It’s
1.A
2.C
3.B
hope it’s correct
Answer:
- <u><em>1.7 × 10³ kg of ore.</em></u>
Explanation:
Call X the amount of aluminum ore mined to produce 1.0 × 10³ kg the aluminum metal.
Then, taking into account the yield of the reaction (82 % = 0.82) and the percent of aluminun in the ore (71% = 0.71), you can write the following equation:
- X × 71% × 82% = 1.0 × 10³ kg
↑ ↑ ↑ ↑
(mass of ore) (% of Al in the ore) (yield) ( Al metal to obtain)
You must just simplify, solve and compute:
- X = 1,000 / (0.71 × 0.82) = 1,000 / 0.5822 = 1,717.6 Kg
Round to two significant figures; 1,700 kg = 1.7 × 10³ kg of ore ← answer.
